Question

ABCD is a trapezium having $AB \parallel DC$. Prove that O, the point of intersection of diagonals, divides the two diagonals in the same ratio. Also prove that $\frac{ar(\triangle OCD)}{ar(\triangle OAB)}= \frac {1}{9}$, if AB = 3 CD.

Answer 1

Answer:

It is proved that ar( ΔOCD)/ar(ΔOAB) = 1/9  

Step-by-step explanation:

Given :  

In trapezium ABCD , AB || DC. AB = 3 CD

(i) Now, in ΔOAB and ΔOCD

∠AOB  = ∠COD    [Vertically opposite angles]

∠OAB   = ∠OCD    [alternate interior angles]

ΔOAB∼ΔOCD  [By AA similarity]

Therefore, OA/OC = OB/OD    

[Since, triangles  are Similar ,hence, corresponding Sides will be proportional]

(ii)  

Since, ΔOAB∼ΔOCD

ar( ΔOCD)/ar(ΔOAB) = (CD/AB)²

[The ratio of area of two similar triangles is equal to the ratio of squares of their corresponding sides.]

ar( ΔOCD)/ar(ΔOAB) = (CD/3 CD)²

[AB = 3 CD]

ar( ΔOCD)/ar(ΔOAB) = CD²/9 CD²

ar( ΔOCD)/ar(ΔOAB) = 1/9  

Hence, it is proved that ar( ΔOCD)/ar(ΔOAB) = 1/9  

HOPE THIS ANSWER WILL HELP YOU ..

Answer 2

Answer:

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